lim n → ∞ 1 + 2 + … + n n 3 2 =

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\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =

MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2019 (21 Apr Shift 2)

Options:

A $0$
B $\frac{2}{3}$
C $1$
D $\frac{3}{2}$

Solution:

1183 Upvotes Verified Answer

The correct answer is:

\frac{2}{3}

The expression is given as,

$I = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots . + \sqrt{n}}{n^{\frac{3}{2}}}$

$= \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots . . + \sqrt{n}}{n \sqrt{n}}$

$= \lim_{n \to \infty} \frac{1}{n} [\sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \dots + \sqrt{\frac{n}{n}}]$

$= \sum_{r = 1}^{n} \lim_{n \to \infty} \frac{1}{n} \sqrt{\frac{1}{n}}$

$= \int_{0}^{1} \sqrt{x} d x$

$= \frac{2}{3} {[x^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3}$

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